dfsb1

Description: Alternate definition of substitution. Remark 9.1 in Megill p. 447 (p. 15 of the preprint). This was the original definition before df-sb . Note that it does not require dummy variables in its definiens; this is done by having x free in the first conjunct and bound in the second. Usage of this theorem is discouraged because it depends on ax-13 . (Contributed by BJ, 9-Jul-2023) Revise df-sb . (Revised by Wolf Lammen, 29-Jul-2023) (New usage is discouraged.)

		Ref	Expression
	Assertion	dfsb1	$⊢ [y/ x] φ \leftrightarrow ((x = y \to φ) \land \exists x (x = y \land φ))$

Proof

Step	Hyp	Ref	Expression
1		sbequ2	$⊢ x = y \to ([y/ x] φ \to φ)$
2	1	com12	$⊢ [y/ x] φ \to (x = y \to φ)$
3		sb1	$⊢ [y/ x] φ \to \exists x (x = y \land φ)$
4	2 3	jca	$⊢ [y/ x] φ \to ((x = y \to φ) \land \exists x (x = y \land φ))$
5		id	$⊢ x = y \to x = y$
6		sbequ1	$⊢ x = y \to (φ \to [y/ x] φ)$
7	5 6	embantd	$⊢ x = y \to ((x = y \to φ) \to [y/ x] φ)$
8	7	sps	$⊢ \forall x x = y \to ((x = y \to φ) \to [y/ x] φ)$
9	8	adantrd	$⊢ \forall x x = y \to (((x = y \to φ) \land \exists x (x = y \land φ)) \to [y/ x] φ)$
10		sb3	$⊢ \neg \forall x x = y \to (\exists x (x = y \land φ) \to [y/ x] φ)$
11	10	adantld	$⊢ \neg \forall x x = y \to (((x = y \to φ) \land \exists x (x = y \land φ)) \to [y/ x] φ)$
12	9 11	pm2.61i	$⊢ ((x = y \to φ) \land \exists x (x = y \land φ)) \to [y/ x] φ$
13	4 12	impbii	$⊢ [y/ x] φ \leftrightarrow ((x = y \to φ) \land \exists x (x = y \land φ))$

Metamath Proof Explorer

Theorem dfsb1

Proof